Solutions to Linear Algebra, Fourth Edition, Stephen H

276 Pages·2011·1.49 MB·English

Solutions to Linear Algebra, Fourth Edition,
Stephen H. Friedberg, Arnold J. Insel,
Lawrence E. Spence
Jephian Lin, Shia Su, Zazastone Lai
July 27, 2011
Copyright © 2011 Chin-Hung Lin. Permission is granted to copy,
distributeand/ormodifythisdocumentunderthetermsoftheGNU
Free Documentation License, Version 1.3 or any later version pub-
lished by the Free Software Foundation; with no Invariant Sections,
no Front-Cover Texts, and no Back-Cover Texts. A copy of the li-
cense is included in the section entitled “GNU Free Documentation
License”.
1
Prologue
ThisisSolutiontoLinearAlgebrawrittenbyFriedberg,Insel,andSpence. And
this file is generated during the Linear Algebra courses in Fall 2010 and Spring
2011. I was a TA in these courses. Although this file will be uploaded to the
coursewebsiteforstudents,themainpurposetowritethesolutionistodosome
exercises and find some ideas about my master thesis, which is related to some
topic in graph theory called the minimum rank problem.
Here are some important things for students and other users. The first is
that there must be several typoes and errors in this file. I would be very glad if
someonesendmeanemail,[email protected],andgivemesomecomments
or corrections. Second, for students, the answers here could not be the answer
on any of your answer sheet of any test. The reason is the answers here are
simplistic and have some error sometimes. So it will not be a good excuse that
your answers is as same as answers here when your scores flied away and leave
you alone.
The file is made by MikTex and Notepad++ while the graphs in this file
is drawn by IPE. Some answers is mostly computed by wxMaxima and little
computed by WolframAlpha. The English vocabulary is taught by Google Dic-
tionary. I appreciate those persons who ever gave me a hand including people
related to those mentioned softwares, those persons who buttressed me, and of
course those instructors who ever taught me.
Thanks.
-Jephian Lin
Department of Mathematrics, National Taiwan University
2011, 5/1
A successful and happy life requires life long hard working.
Prof. Peter Shiue
2
Version Info
• 2011, 7/27—First release with GNU Free Documentation License.
3
Contents
1 Vector Spaces 6
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.2 Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.3 Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.4 Linear Combinations and Systems of Linear Equations . . . . . . 13
1.5 Linear Dependence and Linear Independence . . . . . . . . . . . . 15
1.6 Bases and Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . 17
1.7 Maximal Linearly Independent Subsets. . . . . . . . . . . . . . . . 24
2 Linear Transformations and Matrices 26
2.1 Linear Transformations, Null Spaces, and Ranges . . . . . . . . . 26
2.2 The Matrix Representation of a Linear Transformation . . . . . . 32
2.3 Composition of Linear Transformations and Matrix Multiplication 36
2.4 Invertibility and Isomorphisms . . . . . . . . . . . . . . . . . . . . . 42
2.5 The Change of Coordinate Matrix . . . . . . . . . . . . . . . . . . 47
2.6 Dual Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
2.7 Homogeneous Linear Differential Equations with Constant Coe-
ficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
3 Elementary Matrix Operations and Systems of Linear Equa-
tions 65
3.1 Elementary Matrix Operations and Elementary Matrices . . . . . 65
3.2 The Rank of a Matrix and Matrix Inverses . . . . . . . . . . . . . 69
3.3 Systems of Linear Equation—Theoretical Aspects . . . . . . . . . 76
3.4 Systems of Linear Equations—Computational Aspects . . . . . . 79
4 Determinants 86
4.1 Determinants of Order 2 . . . . . . . . . . . . . . . . . . . . . . . . 86
4.2 Determinants of Order n . . . . . . . . . . . . . . . . . . . . . . . . 89
4.3 Properties of Determinants . . . . . . . . . . . . . . . . . . . . . . . 92
4.4 Summary—Important Facts about Determinants. . . . . . . . . . 100
4.5 A Characterization of the Determinant. . . . . . . . . . . . . . . . 102
4
5 Diagonalization 108
5.1 Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . . . . 108
5.2 Diagonalizability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
5.3 Matrix Limits and Markov Chains . . . . . . . . . . . . . . . . . . 123
5.4 Invariant Subspace and the Cayley-Hamilton Theorem . . . . . . 132
6 Inner Product Spaces 146
6.1 Inner Products and Norms . . . . . . . . . . . . . . . . . . . . . . . 146
6.2 The Gram-Schmidt Orthogonalization Process and Orthogonal
Complements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161
6.3 The Adjoint of a Linear Operator . . . . . . . . . . . . . . . . . . . 170
6.4 Normal and Self-Adjoint Operators . . . . . . . . . . . . . . . . . . 178
6.5 Unitary and Orthogonal Operators and Their Matrices . . . . . . 190
6.6 Orthogonal Projections and the Spectral Theorem . . . . . . . . . 203
6.7 The Singular Value Decomposition and the Pseudoinverse . . . . 208
6.8 Bilinear and Quadratic Forms . . . . . . . . . . . . . . . . . . . . . 218
6.9 Einstein’s Special Theory of Relativity . . . . . . . . . . . . . . . . 228
6.10 Conditioning and the Rayleigh Quotient . . . . . . . . . . . . . . . 231
6.11 The Geometry of Orthogonal Operators . . . . . . . . . . . . . . . 235
7 Canonical Forms 240
7.1 The Jordan Canonical Form I . . . . . . . . . . . . . . . . . . . . . 240
7.2 The Jordan Canonical Form II . . . . . . . . . . . . . . . . . . . . . 245
7.3 The Minimal Polynomial . . . . . . . . . . . . . . . . . . . . . . . . 256
7.4 The Rational Canonical Form . . . . . . . . . . . . . . . . . . . . . 260
GNU Free Documentation License 265
1. APPLICABILITY AND DEFINITIONS . . . . . . . . . . . . . . . . 265
2. VERBATIM COPYING . . . . . . . . . . . . . . . . . . . . . . . . . . 267
3. COPYING IN QUANTITY . . . . . . . . . . . . . . . . . . . . . . . . 267
4. MODIFICATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268
5. COMBINING DOCUMENTS. . . . . . . . . . . . . . . . . . . . . . . 270
6. COLLECTIONS OF DOCUMENTS . . . . . . . . . . . . . . . . . . 270
7. AGGREGATION WITH INDEPENDENT WORKS . . . . . . . . . 270
8. TRANSLATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271
9. TERMINATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271
10. FUTURE REVISIONS OF THIS LICENSE . . . . . . . . . . . . . 272
11. RELICENSING . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272
ADDENDUM: How to use this License for your documents. . . . . . . 273
Appendices 274
5
Chapter 1
Vector Spaces
1.1 Introduction
1. (a) No. 3 ≠ 1
6 4
(b) Yes. −3(−3,1,7)=(9,−3,−21)
(c) No.
(d) No.
2. Here t is in F.
(a) (3,−2,4)+t(−8,9,−3)
(b) (2,4,0)+t(−5,−10,0)
(c) (3,7,2)+t(0,0,−10)
(d) (−2,−1,5)+t(5,10,2)
3. Here s and t are in F.
(a) (2,−5,−1)+s(−2,9,7)+t(−5,12,2)
(b) (3,−6,7)+s(−5,6,−11)+t(2,−3,−9)
(c) (−8,2,0)+s(9,1,0)+t(14,3,0)
(d) (1,1,1)+s(4,4,4)+t(−7,3,1)
4. Additive identity, 0, should be the zero vector, (0,0,...,0) in Rn.
5. Since x = (a ,a )−(0,0) = (a ,a ), we have tx = (ta ,ta ). Hence the
1 2 1 2 1 2
head of that vector will be (0,0)+(ta ,ta )=(ta ,ta ).
1 2 1 2
6. The vector that emanates from (a,b) and terminates at the midpoint
should be 1(c−a,d−b). So the coordinate of the midpoint will be (a,b)+
2
1(c−a,d−b)=((a+c)/2,(b+d)/2).
2
6
7. LetthefourverticesoftheparallelogrambeA,B,C,D counterclockwise.
Say x = A⃗B and y = A⃗D. Then the line joining points B and D should
be x+s(y−x), where s is in F. And the line joining points A and C
should be t(x+y), where t is in F. To find the intersection of the two
lines we should solve s and t such that x+s(y−x)=t(x+y). Hence we
have (1−s−t)x=(t−s)y. But since x and y can not be parallel, we have
1−s−t=0 and t−s=0. So s=t= 1 and the midpoint would be the head
2
of the vector 1(x+y) emanating from A and by the previous exercise we
2
know it’s the midpoint of segment AC or segment BD.
1.2 Vector Spaces
1. (a) Yes. It’s condition (VS 3).
(b) No. If x, y are both zero vectors. Then by condition (VS 3) x =
x+y=y.
(c) No. Let e be the zero vector. We have 1e=2e.
(d) No. It will be false when a=0.
(e) Yes.
(f) No. It has m rows and n columns.
(g) No.
(h) No. For example, we have that x+(−x)=0.
(i) Yes.
(j) Yes.
(k) Yes. That’s the definition.
2. It’s the 3×4 matrix with all entries =0.
3. M =3, M =4, M =5.
13 21 22
6 3 2
4. (a) ( ).
−4 3 9
⎛ 1 −1 ⎞
(b) ⎜ 3 −5 ⎟.
⎝ ⎠
3 8
8 20 −12
(c) ( ).
4 0 28
⎛ 30 −20 ⎞
(d) ⎜ −15 10 ⎟.
⎝ −5 −40 ⎠
(e) 2x4+x3+2x2−2x+10.
(f) −x3+7x2+4.
7
(g) 10x7−30x4+40x2−15x.
(h) 3x5−6x3+12x+6.
⎛ 8 3 1 ⎞ ⎛ 9 1 4 ⎞ ⎛ 17 4 5 ⎞
5. ⎜ 3 0 0 ⎟+⎜ 3 0 0 ⎟=⎜ 6 0 0 ⎟.
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
3 0 0 1 1 0 4 1 0
⎛ 4 2 1 3 ⎞
6. M =⎜ 5 1 1 4 ⎟. Sincealltheentrieshasbeendoubled,wehave2M
⎝ ⎠
3 1 2 6
candescribetheinventoryinJune. Next, thematrix2M−Acandescribe
thelistofsolditems. Andthenumberoftotalsolditemsisthesumofall
entries of 2M −A. It equals 24.
7. It’s enough to check f(0)+g(0)=2=h(0) and f(1)+g(1)=6=h(1).
8. By (VS 7) and (VS 8), we have (a+b)(x+y) = a(x+y)+b(x+y) =
ax+ay+bx+by.
9. Fortwozerovectors0 and0 ,byThm1.1wehavethat0 +x=x=0 +x
0 1 0 1
implies0 =0 ,wherexisanarbitraryvector. Ifforvectorxwehavetwo
0 1
inverse vectors y and y . Then we have that x+y = 0 = x+y implies
0 1 0 1
y =y . Finally we have 0a+1a=(0+1)a=1a=0+1a and so 0a=0.
0 1
10. Wehavesumoftwodifferentiablereal-valuedfunctionsorproductofscalar
andonedifferentiablereal-valuedfunctionareagainthatkindoffunction.
And the function f =0 would be the 0 in a vector space. Of course, here
the field should be the real numbers.
11. All condition is easy to check because there is only one element.
12. Wehavef(−t)+g(−t)=f(t)+g(t)andcf(−t)=cf(t)iff andg areboth
even function. Futhermore, f = 0 is the zero vector. And the field here
should be the real numbers.
13. No. If it’s a vector space, we have 0(a ,a ) = (0,a ) be the zero vector.
1 2 2
But since a is arbitrary, this is a contradiction to the uniqueness of zero
2
vector.
14. Yes. All the condition are preserved when the field is the real numbers.
15. No. Because a real-valued vector scalar multiply with a complex number
will not always be a real-valued vector.
16. Yes. Alltheconditionarepreservedwhenthefieldistherationalnumbers.
17. No. Since 0(a ,a )=(a ,0) is the zero vector but this will make the zero
1 2 1
vector not be unique, it cannot be a vector space.
18. No. We have ((a ,a )+(b ,b ))+(c ,c )=(a +2b +2c ,a +3b +3c )
1 2 1 2 1 2 1 1 1 2 2 2
but (a ,a )+((b ,b )+(c ,c ))=(a +2b +4c ,a +3b +9c ).
1 2 1 2 1 2 1 1 1 2 2 2
8
19. No. Because (c+d)(a ,a )=((c+d)a , a2 ) may not equal to c(a ,a )+
1 2 1 c+d 1 2
d(a ,a )=(ca +dc ,a2 + a2).
1 2 1 1 c d
20. Asequencecanjustbeseenasavectorwithcountable-infinitedimensions.
Or we can just check all the condition carefully.
21. Let 0 and 0 be the zero vector in V and W respectly. Then we have
V W
(0 ,0 ) will be the zero vector in Z. The other condition could also be
V W
checked carefully. This space is called the direct product of V and W.
22. Since each entries could be 1 or 0 and there are m×n entries, there are
2m×n vectors in that space.
1.3 Subspaces
1. (a) No. ThisshouldmakesurethatthefieldandtheoperationsofV and
W arethesame. Otherwiseforexample,V =RandW =Qrespectly.
Then W is a vector space over Q but not a space over R and so not
a subspace of V.
(b) No. We should have that any subspace contains 0.
(c) Yes. We can choose W =0.
(d) No. Let V =R, E ={0} and E ={1}. Then we have E ∩E =∅ is
0 1 0 1
not a subspace.
(e) Yes. Only entries on diagonal could be nonzero.
(f) No. It’s the summation of that.
(g) No. But it’s called isomorphism. That is, they are the same in view
of structure.
−4 5
2. (a) ( ) with tr=−5.
2 −1
⎛ 0 3 ⎞
(b) ⎜ 8 4 ⎟.
⎝ −6 7 ⎠
−3 0 6
(c) ( ).
9 −2 1
⎛ 10 2 −5 ⎞
(d) ⎜ 0 −4 7 ⎟ with tr=12.
⎝ −8 3 6 ⎠
⎛ 1 ⎞
⎜ −1 ⎟
(e) ⎜ ⎟.
⎜ 3 ⎟
⎝ ⎠
5
9